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JEE • MATHEMATICS • LIMITS

Limits practice · Easy

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361 Easy
361 of 647
Analyzing +4 | -1
If ff is a real-valued function satisfying the relation

f(x)+2f(1x)=3xf(x)+2 f\left(\frac{1}{x}\right)=3 x

for all real x≠0x \neq 0, then lim⁡x→0(sin⁡x)f(x)\lim \limits_{{x \rightarrow 0}}(\sin x) f(x) is equal to
View answer and explanation
Correct option: B
2
Option B (2) is correct because by substituting 1/x into the original equation, we can solve for f(x) and then evaluate the limit.
362 Easy
362 of 647
Understanding +4 | -1
The value of lim⁡n→∞(11⋅3+13⋅5+⋯+\lim \limits_{{n \rightarrow \infty}}\left(\frac{1}{1 \cdot 3}+\frac{1}{3 \cdot 5}+\cdots+\right. up to nn terms)) is
View answer and explanation
Correct option: B
1/21 / 2
Option B (1/21 / 2) is correct because the given series is a telescoping series, and its sum converges to 1/2 as n approaches infinity.
363 Easy
363 of 647
Analyzing +4 | -1
Let f(x)={tan⁡2[x]x2−[x]2for x>01∣x∣cot⁡∣x∣for x=0, where [x]for x<0f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\frac{{{{\tan }^2}\left[ x \right]}}{{{x^2} - {{\left[ x \right]}^2}}}}&{{\text{for}}\,x > 0} \\ {\frac{1}{{\sqrt {\left| x \right|\cot \left| x \right|} }}}&\begin{gathered} {\text{for}}\,x = 0,\, \quad \\ {\text{where }}\left[ x \right] \quad \\ {\text{for}}\,x < 0 \quad \\ \end{gathered} \end{array}} \right.
and {x} denote respectively the greatest integer less than equal to xx and the fractional part of xx, then
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because the left-hand limit, right-hand limit, and the cotangent inverse of the squared left-hand limit all satisfy the given conditions.
364 Easy
364 of 647
Application +4 | -1
The value of lim⁡x→ax−b−a−bx2−a2(a>b)\lim \limits_{{x \rightarrow a}} \frac{\sqrt{x-b}-\sqrt{a-b}}{x^2-a^2}(a>b) is
View answer and explanation
Correct option: D
14aa−b\frac{1}{4 a \sqrt{a-b}}
Option D (14aa−b\frac{1}{4 a \sqrt{a-b}}) is correct because after rationalizing and simplifying, the limit evaluates to 14aa−b\frac{1}{4a\sqrt{a-b}}.
365 Easy
365 of 647
Application +4 | -1
lim⁡x→0cos⁡2x−cos⁡xsin⁡2x\lim \limits_{{x \rightarrow 0}} \frac{\cos 2 x-\cos x}{\sin ^2 x} is
View answer and explanation
Correct option: D
−32-\frac{3}{2}
The correct answer is D (-32\frac{3}{2}) because applying L'Hôpital's rule twice to the limit yields the value -32\frac{3}{2}.
366 Easy
366 of 647
Application +4 | -1
lim⁡x→0(1−ax)1/x\lim\limits_{x \to 0} {\left( {1 - ax} \right)^{1/x}} is equal to
View answer and explanation
Correct option: A
e−a{e^{ - a}}
Option A (e−a{e^{-a}}) is correct because it is the direct application of the standard limit formula for exponential functions.
367 Easy
367 of 647
Analyzing +4 | -1
The value of lim⁡x→∞x3/2(x3+1−x3−1)\lim\limits_{x \to \infty } {x^{3/2}}\left( {\sqrt {{x^3} + 1} - \sqrt {{x^3} - 1} } \right), is
View answer and explanation
Correct option: A
1
Option A (1) is correct because after rationalizing and simplifying, the limit evaluates to 1.
368 Easy
368 of 647
Application +4 | -1
The value of cc for which the conclusion of Lagrange's theorem holds for the function f(x)=f(x)= a2−x2,a>1\sqrt{a^2-x^2}, a>1 on the interval [1,a][1, \mathrm{a}] is
View answer and explanation
Correct option: C
a(a+1)2\frac{\sqrt{a(a+1)}}{2}
Option C (a(a+1)2\frac{\sqrt{a(a+1)}}{2}) is correct because the value of c that satisfies Lagrange's theorem for the given function and interval is a(a+1)2\frac{\sqrt{a(a+1)}}{2}, found by equating the derivative at c to the average rate of change.
369 Easy
369 of 647
Understanding +4 | -1
If f(x)=∫1xsin⁡ttdtf(x)=\int\limits_{1}^x \frac{\sin t}{t} d t, then lim⁡x→∞f′(x)\lim \limits_{{x \rightarrow \infty}} f^{\prime}(x) is equal to
View answer and explanation
Correct option: A
0
Option A (0) is correct because as x approaches infinity, the function f'(x) = sin⁡tt\frac{{\sin t}}{{t}} approaches 0, due to the denominator growing without bound while the numerator oscillates between -1 and 1.
370 Easy
370 of 647
Application +4 | -1
lim⁡x→01−cos⁡2xx=\lim \limits_{{x \rightarrow 0}} \frac{1-\cos 2 x}{x}=
View answer and explanation
Correct option: A
0
Option A (0) is correct because lim⁡x→01−cos⁡2xx=0\lim \limits_{{x \rightarrow 0}} \frac{1-\cos 2 x}{x} = 0.
371 Easy
371 of 647
Analyzing +4 | -1
If lim⁡x→0aex−bcos⁡x+ce−xxsin⁡x=2\lim \limits_{{x \rightarrow 0}} \frac{a e^x-b \cos x+c e^{-x}}{x \sin x}=2, then a+b+c=a+b+c=
View answer and explanation
Correct option: B
4
Option B (4) is correct because the limit exists only when certain conditions are met, which leads to a system of equations that can be solved for a+b+ca+b+c.
372 Easy
372 of 647
Application +4 | -1
lim⁡n→∞12+22+32+...+n2n3\lim\limits_{n \to \infty } \frac{{{1^2} + {2^2} + {3^2} + ... + {n^2}}}{{{n^3}}} is equal to
View answer and explanation
Correct option: C
13\frac{1}{3}
Option C (13\frac{1}{3}) is correct because the limit of the sum of squares divided by n3 is 1/3.
373 Easy
373 of 647
Analyzing +4 | -1
 Let f(θ)=1tan⁡2θ{(1+tan⁡θ)3+(2+tan⁡θ)3+….+(10+tan⁡θ)3}−10tan⁡θ\begin{aligned}\text { Let } f(\theta)=\frac{1}{\tan ^2 \theta}\left\{(1+\tan \theta)^3+(2+\tan \theta)^3+\ldots\right..\left.+(10+\tan \theta)^3\right\}-10 \tan \theta\end{aligned}

Then, lim⁡θ→π−2f(θ)\lim \limits_{\theta \rightarrow \frac{\pi^{-}}{2}} f(\theta) is equal to
View answer and explanation
Correct option: C
165
Option C (165) is correct because the limit is evaluated by simplifying the expression and applying limit properties as θ approaches π−2\frac{π^{-}}{2}.
374 Easy
374 of 647
Analyzing +4 | -1
lim⁡x→∞2+2x+sin⁡2x(2x+sin⁡2x)esin⁡x\lim \limits_{{x \rightarrow \infty}} \frac{2+2 x+\sin 2 x}{(2 x+\sin 2 x) e^{\sin x}} is equal to
View answer and explanation
Correct option: D
Non-existent
Option D (Non-existent) is correct because the limit oscillates between 1e\frac{1}{e} and e as x approaches infinity due to the esin⁡xe^{\sin x} term.
375 Easy
375 of 647
Analyzing +4 | -1
If [.] denotes the greatest integer function, then lim⁡x→0tan⁡([−2π2]x2)−x2tan⁡[−2π2]sin⁡2x\lim \limits_{{x \rightarrow 0}} \frac{\tan \left(\left[-2 \pi^2\right] x^2\right)-x^2 \tan \left[-2 \pi^2\right]}{\sin ^2 x} is equal to
View answer and explanation
Correct option: A
−20+tan⁡20-20+\tan 20
Option A (−20+tan⁡20-20+\tan 20) is correct because it is the result of applying limit rules and trigonometric identities to the given expression.

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376 Easy
376 of 647
Analyzing +4 | -1
The value of lim⁡x→∞1+x4−(1+x2)x2\lim\limits_{x \to \infty } \frac{{\sqrt {1 + {x^4}} - \left( {1 + {x^2}} \right)}}{{{x^2}}} is equal to
View answer and explanation
Correct option: A
0
Option A (0) is correct because after dividing numerator and denominator by x2 and taking the limit, the expression evaluates to 0.
377 Easy
377 of 647
Application +4 | -1
lim⁡x→+∞x−83−x+10x2=\lim _{x \rightarrow+\infty} \frac{x-8}{3-x+10 x^2}=
View answer and explanation
Correct option: A
0
Option A (0) is correct because after dividing by x2 and simplifying, the limit of the expression as x approaches infinity is 0.
378 Easy
378 of 647
Application +4 | -1
Value of lim⁡x→01+tan⁡x3−1−tan⁡x3x\lim \limits_{{x \rightarrow 0}} \frac{\sqrt[3]{1+\tan x}-\sqrt[3]{1-\tan x}}{x} is-
View answer and explanation
Correct option: D
23\frac{2}{3}
Option D (2/3) is correct because the limit is evaluated using L'Hopital's rule.
379 Easy
379 of 647
Analyzing +4 | -1
If l=lim⁡x→0tan⁡xn(tan⁡x)ml=\lim \limits_{{x \rightarrow 0}} \frac{\tan x^n}{(\tan x)^m}, where m,n∈Nm, n \in N, then
View answer and explanation
Correct option: C
l={1, if n=m0, if n>ml= \begin{cases}1, \text { if } n=m \\ 0, \text { if } n > m\end{cases}
The correct answer is l={1, if n=m0, if n>ml= \begin{cases}1, \text { if } n=m \\ 0, \text { if } n > m\end{cases} because the limit depends on the relationship between nn and mm. When n=mn=m, the limit is 1, and when n>mn>m, the limit is 0.
380 Easy
380 of 647
Analyzing +4 | -1
If f(x)={tan⁡−1(x+[x])[x]−2x,[x]≠00,[x]=0f(x)=\left\{\begin{array}{cl}\frac{\tan ^{-1}(x+[x])}{[x]-2 x}, {[x] \neq 0} \\ 0,[x]=0\end{array}\right. where [x][x] denotes the greatest integer less than or equal to xx, then lim⁡x→0f(x)\lim \limits_{{x \rightarrow 0}} f(x) is
View answer and explanation
Correct option: D
Non-existent
Option D (Non-existent) is correct because the left-hand limit and right-hand limit are not equal, hence the limit does not exist.
381 Easy
381 of 647
Application +4 | -1
Find lim⁡x→0(x+9)32+ex−28x\lim \limits_{{x \rightarrow 0}} \frac{(x+9)^{\frac{3}{2}}+e^x-28}{x}
View answer and explanation
Correct option: D
11/211 / 2
Option D (11/211 / 2) is correct because the limit can be evaluated by recognizing it as the derivative of the function at x=0.
382 Easy
382 of 647
Analyzing +4 | -1
The value of lim⁡n→∞{n2−n33+n}\lim \limits_{{n \rightarrow \infty}}\left\{\sqrt[3]{n^2-n^3}+n\right\}, is
View answer and explanation
Correct option: A
13\frac{1}{3}
Option A (1/3) is correct because, after rationalizing and simplifying, the limit evaluates to 1/3.
383 Easy
383 of 647
Application +4 | -1
If f(a)=2,f′(a)=1,g(a)=3,g′(a)=−1f\left( a \right) = 2,f'\left( a \right) = 1,g\left( a \right) = 3,g'\left( a \right) = - 1, then lim⁡x→af(a)g(x)−f(x)g(a)x−a\lim\limits_{x \to a} \frac{{f\left( a \right)g\left( x \right) - f\left( x \right)g\left( a \right)}}{{x - a}} is equal to
View answer and explanation
Correct option: D
−5- 5
Option D (-5) is correct because applying L'Hopital's rule and substituting the given values results in -5.
384 Easy
384 of 647
Application +4 | -1
The value of lim⁡x→1y→0y3x3−y2−1\lim \limits_{\substack{x \rightarrow 1 \\ y \rightarrow 0}} \frac{y^3}{x^3-y^2-1} as (x,y)→(1,0)(x, y) \rightarrow(1,0) along the line y=x−1y=x-1 is-
View answer and explanation
Correct option: A
0
Option A (0) is correct because substituting y = x - 1 and taking the limit as y approaches 0 results in a value of 0.
385 Easy
385 of 647
Analyzing +4 | -1
If f(x)=lim⁡y→x  sin⁡2y  −  sin⁡2  xy2  −  x2f\left( x \right) = \lim\limits_{y \to x} \,\,\frac{{{{\sin }^2}y\,\, - \,\,{{\sin }^2}\,\,x}}{{{y^2}\,\, - \,\,{x^2}}}, then ∫4x    f(x)  dx\int 4x\,\,\;f\left( x \right)\,\,dx is equal to
View answer and explanation
Correct option: C
− cos⁡  2x  +  c- \,\cos \,\,2x\,\, + \,\,c
Option C is correct because after applying L'Hôpital's rule and integrating, the result is − cos⁡ 2x  +  c- \,\cos \,2x\,\, + \,\,c.
386 Easy
386 of 647
Analyzing +4 | -1
Let f(a)=g(a)=kf\left( a \right) = g\left( a \right) = k and their nth derivatives fn(a),gn(a){f^n}\left( a \right),{g^n}\left( a \right) exist and are not equal for somen. Further if lim⁡x→af(a)g(x)−f(a)−g(a)f(x)+g(a)g(x)−f(x)=4\lim\limits_{x \to a} \frac{{f\left( a \right)g\left( x \right) - f\left( a \right) - g\left( a \right)f\left( x \right) + g\left( a \right)}}{{g\left( x \right) - f\left( x \right)}} = 4, then the value of kk is equal to
View answer and explanation
Correct option: A
4
Option A (4) is correct because by applying L'Hôpital's rule, the limit simplifies to k, which is equal to 4.
387 Easy
387 of 647
Application +4 | -1
lim⁡x→02sin⁡23xx2\lim\limits_{x \to 0} \frac{{2{{\sin }^2}3x}}{{{x^2}}} is equal to
View answer and explanation
Correct option: C
18
Option C (18) is correct because by using the limit lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{{\sin x}}{x} = 1, the given limit simplifies to 18.
388 Easy
388 of 647
Application +4 | -1
Let f(x)={sin⁡4xlog⁡(1+3x),x≠0A+1,x=0f(x)=\left\{\begin{array}{cc}\frac{\sin 4 x}{\log (1+3 x)}, & x \neq 0 \\ A+1, & x=0\end{array}\right.

The value of AA for ff to be continuous at x=0x=0 is
View answer and explanation
Correct option: B
13\frac{1}{3}
Option B (13\frac{1}{3}) is correct because for ff to be continuous at x=0x=0, A+1=lim⁡x→0sin⁡4xlog⁡(1+3x)=43A+1 = \lim_{x \rightarrow 0} \frac{\sin 4x}{\log(1+3x)} = \frac{4}{3}, which gives A=13A = \frac{1}{3}.
389 Easy
389 of 647
Analyzing +4 | -1
lim⁡x→02sin⁡x−sin⁡2xtan⁡3x\lim \limits_{{x \rightarrow 0}} \frac{2 \sin x-\sin 2 x}{\tan ^3 x} is equal to
View answer and explanation
Correct option: B
1
Option B (1) is correct because after simplifying the expression using trigonometric identities, the limit evaluates to 1.
390 Easy
390 of 647
Analyzing +4 | -1
The value of lim⁡x→1{xn−1n(x−1)}1x−1\lim \limits_{{x \rightarrow 1}}\left\{\frac{x^n-1}{n(x-1)}\right\}^{\frac{1}{x-1}}, is
View answer and explanation
Correct option: C
en−12e^{\frac{n-1}{2}}
Option C (en−12e^{\frac{n-1}{2}}) is correct because it is obtained by simplifying the expression using algebraic manipulation and then applying limit properties.
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